Question

Given sec θ = `13/12`, calculate all other trigonometric ratios.

Answer 1

Let ΔABC be a right-angled triangle, right angled at point B.

It is given that:

sec θ = `"hypotenuse"/"side adjacent to ∠θ" = "AC"/"AB" = 13/12`

Let AC = 13k and AB = 12k, where k is a positive integer.

Applying pythagoras theorem in Δ ABC, we obtain:

AC² = AB² + BC²

BC² = AC² - AB²

BC² = (13k)² - (12k)²

BC² = 169 k² - 144 k²

BC² = 25k²

BC = 5k

sin θ = `("side opposite to ∠θ")/("hypotenuse") = ("BC")/("AC") = 5/13`

cos θ = `("side adjacent to ∠θ")/("hypotenuse") = ("AB")/("AC") = 12/13`

tan θ = `("side opposite to ∠θ")/("side adjacent to ∠θ") = "(BC)"/"(AB)" = 5/12`

cot θ = `("side adjacent to ∠θ")/("side opposite to ∠θ") = ("AB")/("BC") = 12/5`

cosec θ = `("hypotenuse")/("side opposite to ∠θ") = ("AC")/("BC") = 13/5`